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3 April, 07:06

For a particular lab experiment, you need 150 mL of a 2.50 M solution of ammonium nitrate. How many grams of NH4NO3 will you need for the solution?

8.0 g

10.02 g

25.0 g

875.0 g

30.0 g

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  1. 3 April, 07:31
    0
    mass NH4NO3 = 30.0 g

    Explanation:

    ∴ V = 150 mL

    ∴ C NH4NO3 = 2.50 M

    ⇒ mass NH4NO3 = ?

    ∴ mol NH4NO3 = (0.150 L) * (2.50 mol/L) = 0.375 mol NH4NO3

    ∴ Mw NH4NO3 = 80.043 g/mol

    ⇒ mass NH4NO3 = (0.375 mol) * (80.043 g/mol) = 30.016 g
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